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The sequence \[x _{1},x _{2},x _{3},...\]is defined by \[x _{k}=1/(k ^{2}+k)\]. A sum of consecutive terms \[x _{m}, x _{m+1}, +...+x _{n}=1/29\]. Find n and m.
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I've got this equation for m and n: \[m=\frac{29(n+1)}{n+30}\] We need to find positive integer solutions for n and m.
Obviously \(n>m\).
Notice that \[ \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} \] Hence \[ 1 - \frac{1}{n+1} = \sum_{k=1}^n \frac{1}{k(k+1)} =: S_n\] Now, as 29 is prime, this would seem to suggest we need to write \[ \frac{1}{29} = 1 - \left(1 - \frac{1}{28+1}\right) = S_\infty - S_{28} \]
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